Question

The manager of a small park randomly surveyed residents in the community and asked whether they visited the park in the past year. The table shows the results.

What is the 95% confidence interval for the difference between the percent of residents who are younger than 50 and who are 50 and older that visited the park in the past year?

Answer 1

@ganeshie8

Answer 2

is this graph but with different numbers

Answer 3

http://assets.openstudy.com/updates/attachments/554a32b1e4b048e824fed8f4-garcia101-1430926013925-q8_1.jpg

Answer 4

but idk how to change the number in open study

Answer 5

135..............43
92................48

Answer 6

@zepdrix

Answer 7

what is 135...43, 92...48?

Answer 8

the numbers that are supposed to be on the chart but idk how to change them

Answer 9

ok, now how do you usually construct such a confidence interval?

Answer 10

using the difference formula?

Answer 11

i dont know the name of it, but that does sound like an apt description.

something like
\[(p_1-p_2)\pm z\left(\sqrt{\frac{p_1q_1}{n_1^3}+\frac{p_2q_2}{n_2^3}}\right)\]

Answer 12

http://www.kean.edu/~fosborne/bstat/06d2pop.html

Answer 13

yeah, thats the one

Answer 14

so, where is the difficulty ?

Answer 15

how to find the p^ for the formula

Answer 16

with the numbers from the chart

Answer 17

number of people in a group, divided by number of old people in all

Answer 18

p q
135...43 ... (135-43)
92...48 ... ( 92-48)
----
227 people in total

is what im thinking ... but the question seems odd to me. is it asked correctly? no mistakes?

Answer 19

proportions?

Answer 20

and no no mistakes

Answer 21

135 out of 227 people is the proportion that are younger than 50
92 out of 227 people is the proportion that are 50 and older

would you agree?

Answer 22

yes

Answer 23

what next?

Answer 24

still trying to understand the question ....


# surveyed #yes
young 135 43
old 92 48
total 227 91


What is the 95% confidence interval for the difference between the percent of residents who are younger than 50 and who are 50 and older that visited the park in the past year?


(135-92) +- 1.96 sqrt[ 135*92 /227]

seems about the only sensible approach to me, but im unsure about it

Answer 25

a similar question was asked about a year ago if the time stamp is correct

http://openstudy.com/study#/updates/537a329fe4b0e54307961a83

Answer 26

but no one replied

Answer 27

i see that, but the options presented give us a way to hopefully test the approach.

for their work i get 70.96 as an upper limit to the interval, and none of their options has that ....

Answer 28

I have choices too

Answer 29

would it help?

Answer 30

it would help to see of the method used is at least part of the expected process yes

Answer 31

14.9, 25.7
7.4, 33.2
13.7, 29.6
9.5, 31.1

Answer 32

there has to be a different approach then, my idea isnt getting us close to an option.

Answer 33

yeah and I've done a lot of stuff to that formula and nothing seems to work :/

Answer 34

the reason the question seems odd to me is that we are comparing old to young ... but are given how many say yes in each case. and the question is not asking anything about the yes/no portion of it, which would make better sense to me.

Answer 35

EXACTLY thats what im noticing that there is no way to use the survey

Answer 36

3 options give us an average of 20.3 so the correct answer most likely is:

20.3 +- 1.96(S)

Answer 37

but that is not on the choices i gave you

Answer 38

it is at best 3 of the choices, whatever the method is spose to be, we most likely need that left part to come out to 20.3 since that is the middle of 3 of the intervals in the options.

Answer 39

I got lost there

Answer 40

given the 4 options, the average (the middle of the interval) is 20.3 for 3 options.

its safe to say that they expect you to get the part of the setup ... and mess up on the 1.96(S) part of it

M +- 1.96(S) gives us an interval centered around M

in the case of your options, M=20.3 for 3 cases telling us that that portion of the setup is most likely 20.3

Answer 41

im confused im sorry because that makes no sense with the options that I have

Answer 42

options you gave:

14.9, 25.7 ; midpoint is 20.3
7.4, 33.2 ; midpoint is 20.3
13.7, 29.6 ; midpoint is 21.65
9.5, 31.1 ; midpoint is 20.3

Answer 43

ohhhhhhhhok got it

Answer 44

a confidence interval will be centered about a point ... M, and then we measure the number of deviations away from M that comprise 95% of the data.

Answer 45

therefore we have to find a formula with the midpoint?

Answer 46

yes, is there a way to test an option? like do you get a certain number of tries at it?

Answer 47

no just one try :(

Answer 48

if noone else has a better idea, ill do some reading up on this to see if i can make better sense of it. might take a while tho.

Answer 49

is alright :) would it help if I gave you the link to my book?

Answer 50

you have a book? lol, yes it would be helpful

20.3 is a percentage ... decimal is .203

43/135 - 48/92 = .203...

Answer 51

so it appears that the question being asked is about the difference between those who said yes ...

Answer 52

|43/135 - 48/92| + 1.96sqrt(43(135-43)/135^3+48(92-48)/92^3) gives an upper value of .332, or 33.2

Answer 53

the way the question is worded is just horrible

Answer 54

so option B then? according to your calculations

Answer 55

B is the only one that makes any sense to me as far as an answer goes yes

Answer 56

alright thank you very much :) I appreciate your help

Answer 57

had to work backwards ...

the intervals are in percent format, 20.3% is our most likely midpoint from the options

.203 is the decimal format ... we can obtain that from subtracting the yes/age proportions, which means that our standard error calculations for the sqrt part uses the complements of those proportions.

as such, .332 or 33.2% is the high end of our 95% interval (z=1.96)

Answer 58

is it correct? i have no idea lol .... but it is the only one that makes sense to me as far as the options go. the wording of the question itself remains a mystery to me.

Answer 59

it was correct thank you for your help :)

Answer 60

yay!! :) good luck